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Basis Collapse in Holographic Algorithms Over All Domain Sizes Sitan Chen Harvard College March 29, 2016 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof


  1. Basis Collapse in Holographic Algorithms Over All Domain Sizes Sitan Chen Harvard College March 29, 2016

  2. Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k � = 2 K ? Next Steps Acknowledgments

  3. Holographic algorithms reduce counting problems into the problem of counting perfect matchings in a graph G = ( V, E ). • Perfect matching: M ⊂ E for which every v ∈ V belongs to exactly one edge e ∈ M • [Valiant ’79]: Counting perfect matchings in arbitrary graphs is # P -complete. • [Fisher-Temperley 1961, Kasteleyn 1961]: Counting perfect matchings in planar graphs is in P .

  4. More generally, if every edge e of G has some weight w ( e ), define � � � � PerfMatch( G ) = w ( e ) . perfect matchings M e ∈ M Theorem (FKT algorithm) If G is a planar weighted graph, PerfMatch( G ) can be computed in polynomial time. Idea. For an arbitrary graph G with adjacency matrix A , the Pfaffian � � � � Pf( A ) = sgn( M ) w ( e ) perfect matchings M e ∈ M satisfies Pf( A ) 2 = det( A ). For planar graphs, can flip the signs of some entries of A to make Pf and PerfMatch agree.

  5. ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 1 ∨ x 3 ∨ x 4 ) ∧ ( x 5 ∨ x 6 ∨ x 7 ) ∧ ( x 4 ∨ x 5 ∨ x 6 )

  6. ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 1 ∨ x 3 ∨ x 4 ) ∧ ( x 5 ∨ x 6 ∨ x 7 ) ∧ ( x 4 ∨ x 5 ∨ x 6 )

  7. Imagine: each vertex v on the left propagates signals along its outgoing edges indicating whether v is assigned 1 (green) or 0 (black).

  8. Each satisfying assignment corresponds to a collection of signals satisfying two constraints: Satisfaction : If C j is a vertex Consistency : If x i is a vertex on the right, at least one of the on the left, the two signals x i three signals it receives must be generates must be the same. 1. 000 0 001 1 00 1 010 1 01 0 011 1 10 0 100 1 11 1 101 1 110 1 111 1

  9. Goal: encode these bit vectors using the matching properties of graphs Definition A matchgate is a weighted graph G with designated subsets of its vertices called external nodes X . We say that it is of arity | X | . Definition The standard signature G of matchgate G of arity n is a vector of dimension 2 n with entries indexed by bitstrings of length n . For Z ⊂ X corresponding to bitstring α , Γ α = PerfMatch(Γ \ Z ) .

  10. 00 3 01 0 10 0 11 5

  11. 000 0 001 3 010 3 011 0 100 3 101 0 110 0 111 5

  12. We want planar matchgates G and R whose standard signatures respectively match the vectors encoding the consistency and satisfaction constraints: Satisfaction : If C j is a vertex Consistency : If x i is a vertex on the right, at least one of the on the left, the two signals x i three signals it receives must be generates must be the same. 1. 000 0 001 1 00 1 010 1 01 0 011 1 10 0 100 1 11 1 101 1 110 1 111 1

  13. ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 1 ∨ x 3 ∨ x 4 ) ∧ ( x 5 ∨ x 6 ∨ x 7 ) ∧ ( x 4 ∨ x 5 ∨ x 6 )

  14. ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 1 ∨ x 3 ∨ x 4 ) ∧ ( x 5 ∨ x 6 ∨ x 7 ) ∧ ( x 4 ∨ x 5 ∨ x 6 )

  15. ( x 1 ∨ x 2 ∨ x 3 ) ∧ ( x 1 ∨ x 3 ∨ x 4 ) ∧ ( x 5 ∨ x 6 ∨ x 7 ) ∧ ( x 4 ∨ x 5 ∨ x 6 )

  16. Unfortunately, no recognizer has standard signature (0 , 1 , 1 , 1 , 1 , 1 , 1 , 1): Observation (Parity Condition) Because a graph with an odd number of vertices has no perfect matchings, given any matchgate G , the indices of the nonzero entries in its standard signature must have the same parity.

  17. • The saving grace: rewrite number of perfect matchings of matchgrid Ω as an inner product and apply a change of basis. • Suppose there are w wires in Ω, generators G 1 , ...G g , and R 1 , ..., R r recognizers, then   g r � � � G iy i R jx j  = � G , R � , PerfMatch(Ω) =  z ∈{ 0 , 1 } w , i =1 j =1 z = x 1 ◦···◦ x r ◦ y 1 ◦···◦ y g where G = ⊗ i G i and R = ⊗ i R i with the order of tensoring specified by the wires. • Regard G as an element in X = C 2 w and R as an element in X ∗ : PerfMatch(Ω) is the result of applying dual vector R to G , which is independent of the choice of basis for X .

  18. Definition Given a 2 × 2 basis matrix M , the signature with respect to M of a generator G of arity n is the vector G satisfying G = M ⊗ n G. The signature with respect to M of a recognizer R of arity n is the vector R satisfying R = RM ⊗ n .

  19. • Suffices to find a basis M of matchgates G and R whose signatures with respect to M match the vectors encoding the consistency and satisfaction constraints. • Over C and F 2 , this still cannot be done. � 1 � 3 • [Valiant ’06, Cai-Lu ’07]: Over F 7 , take M = , 6 5 G = (3 , 0 , 0 , 5), and R = (0 , 3 , 3 , 0 , 3 , 0 , 0 , 5).

  20. 000 0 00 3 001 3 01 0 010 3 10 0 011 0 11 5 100 3 101 0 110 0 111 5

  21. Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k � = 2 K ? Next Steps Acknowledgments

  22. • The number of different values that objects in a counting problem can take on is called the domain size. • Domain size 2: ◮ Boolean satisfying assignments ◮ Vertex covers ◮ Perfect matchings ◮ Ice problems • Domain size k ◮ k -colorings

  23. Over domain size k : • Arity- n signatures are now vectors of dimension k n . • M now has width k because G = M ⊗ n G R = RM ⊗ n .

  24. • Domain size 2: encode True / False by presence/absence of one external node • Domain size k : encode colors { 1 , ..., k } by removal of some subset of a group of ℓ external nodes ◮ Arities are now multiples of ℓ ◮ External nodes grouped into blocks of ℓ , with wires connecting matchgates blockwise. ◮ If Γ has n blocks, Γ has 2 ℓn entries. ◮ M has height 2 ℓ because G = M ⊗ n G R = RM ⊗ n . ◮ We call ℓ the basis size.

  25. Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k � = 2 K ? Next Steps Acknowledgments

  26. We will regard standard signatures as matrices: Definition For standard signature G of generator G , the t -th matrix form G ( t ) (1 ≤ t ≤ n ) is the 2 ℓ × 2 ( n − 1) ℓ matrix of entries of G where the rows are indexed by α t ∈ { 0 , 1 } ℓ and the columns are indexed by α 1 · · · α t − 1 α t +1 · · · α n ∈ { 0 , 1 } ( n − 1) ℓ .

  27. We will also regard signatures as matrices: Definition For signature G of generator G , the t -th matrix form G ( t ) (1 ≤ t ≤ n ) is the k × k n − 1 matrix of entries of G where the rows are indexed by α t ∈ [ k ] and the columns are indexed by α 1 · · · α t − 1 α t +1 · · · α n ∈ [ k ] n − 1 . Note: we will denote row indices by superscripts and column indices by subscripts.

  28. Definition A generator G is full rank if there exists t for which rank( G ( t )) = k . It turns out we may assume that rank( M ) = k . But we know G ( t ) = MG ( t )( M T ) ⊗ ( n − 1) . So if G is of full rank, rank( G ( t )) = k.

  29. Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size: Question Given k , what is the smallest ℓ for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size ℓ ? domain size basis size Cai-Lu ’08 2 1

  30. Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size: Question Given k , what is the smallest ℓ for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size ℓ ? domain size basis size Cai-Lu ’08 2 1 Cai-Fu ’14 3 1 4 2

  31. Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size: Question Given k , what is the smallest ℓ for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size ℓ ? domain size basis size Cai-Lu ’08 2 1 Cai-Fu ’14 3 1 4 2 C ’15, Xia ’15 k ⌊ log 2 k ⌋

  32. Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k � = 2 K ? Next Steps Acknowledgments

  33. Definition Z ⊂ { 0 , 1 } n is a cluster if there exists s ∈ { 0 , 1 } n and positions p 1 , ..., p m ∈ [ n ] such that each member of Z is of the form �� � s ⊕ j ∈ J e p j for some J ⊂ { p 1 , ..., p m } , where e p j is the bitstring consisting of zeroes everywhere except position p j . We write Z as s + { e p 1 , ..., e p m } ( s only unique up to the bits outside of positions p 1 , ..., p m ). e.g. { 000 , 001 , 100 , 101 } is a cluster denoted 000 + { e 1 , e 3 } .

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